Antiparticles and C, P, and T Transformations (The Standard Model Part 2)
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- Опубліковано 17 тра 2024
- Before we start adding more particles to the standard model, we have to address an elephant in the room. When we try to make special relativity consistent with quantum mechanics, we find that some solutions to the theory don't seem to make sense. In this video, we will show how these can actually be interpreted as additional particles (antiparticles) which we must always take into account when adding new particles to the standard model!
Part 1 here: • Symmetry and Quantum E...
OMG, your channel is seriously amazing, I have never found someone who explains these things SO simply, yet still includes so much of the math. great job, please keep doing what you're doing.
Just found this channel, really helpful videos so far! I sometimes struggle to see the bigger picture when reading textbooks, this helps taking a step back.
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Nice. I wish I had this primer when having the course of QM back at the university.
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Awesome video!
So if I have a system that has charge and parity symmetry, I can't imply any conservation laws?
damn.
Best channel.
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Were are you from? I need answers!
Noether's theorem only applies to continuous stuff? Thanks for mentioning that. I definitely should read more.
Because the theorem contains derivatives , and one cannot derivate non-continuous function
Very nice video!
So when Hawking came up with his particle/anti-particle pair analogy for Hawking Radiation, did he just keep the negative energy interpretation for the anti particle to make the black hole lose mass? That would make that analogy make much more sense. I understand that in modern descriptions there is an ingoing "negative energy partner wave" for the outgoing radiation, but i have not been able to find a source(that i was able to understand...) that explains where that comes from other than just assuming that energy is conserved.
Aren't there still conserved quantities associated with these symmetries, though? Or are those just not a result of some form of Noether's Theorem?
I don't know the answer to your question, but since you mentioned Hawking radiation, I want to make sure you know about a recent video on that subject by a channel called The Science Asylum. Basically, he disagrees with the two-particle analogy.
Hi Narf, glad you enjoyed the video! Luckily, I think the answer to your first question is quite straightforward. It actually doesn't matter that you necessarily have particles versus antiparticles being emitted by the black hole. The important thing is that you are emitting particles at all. Remember, that even when we create an antiparticle, it still has a positive energy (that is the whole point of introducing them to begin with!), so as the black hole radiates, you now have particles where you didn't have particles before, but that energy can't come from nowhere! So where does it come from? The only place that it possibly can come from is the gravitational field since this is the only source of energy available. So, when you radiate (create) a particle, that particle's energy is "stolen" from the gravitational field of the black hole, meaning that the gravitational field now has less energy, which is of course the same statement as saying that the black hole has lost mass.
As for conserved quantities, there does tend to be some overlap especially with CP symmetry and conserved quantities. However, as far as I am aware, the conserved quantities always come from some continuous (usually global) symmetries which are present in many theories which display these discrete symmetries. For example, in QED with only electrons, positrons, and photons, this theory is C, P, and T symmetric, but it also has an extra symmetry under global phase transformations of the electron and positron fields, as long as the positrons transform in the opposite way of the electrons. Here, the conserved charge is what we might call "electron number" where each electron gets a value of +1 and each positron gets a value of -1. So, you see in this case, the two sort of overlap: the discrete symmetries as well as this global charge from the continuous symmetry both ensure that particles and antiparticles are created in equal amounts.
This doesn't have to be the case, though. In the standard model, CP symmetry is violated, but similar conserved charges, like baryon number minus lepton number, still exist, corresponding to global, continuous transformations. Similarly, we can have a CP symmetric theory with no such conserved charges. This is the case for many theories with only interacting, real bosons, where the interactions don't arise from a gauge symmetry. In these cases, the theory doesn't have to respect any sort of "boson number symmetry" so there is no analogous conserved charges, even if the theory is CP symmetric.
Hopefully that answers your questions!
@@zapphysics To be honest, I was hoping for something along the lines of "you see we can construct the QFT vacuum from positive and negative frequency modes which cancel out, but the black hole causes some of these modes to not cancel out, resulting in a far away observer seeing real, positive energy particles while the black hole sees real negative energy particles because math" or something. But it seems there is currently just no actual local "mechanism" that gives something like that?
If we say the energy for the radiation comes from the curvature field and that means the black hole shrinks, that means we define the black hole's mass completely via the field around it, right? So since the curvature changes, so does the mass.
Does that work for the charge as well? Can we define the black hole's charge via it's electric field and if the radiation is in form of charged particles, that charge comes from the electric field and the black hole's charge has changed?
As usual thank you a lot for your answers!
Also, sorry if this is getting too far away from the scope of this video.
@@narfwhals7843 The calculation of Hawking radiation itself actually looks nothing like Hawking's "layman" explanation of particle/antiparticle pairs being created near the horizon and one of the pairs escaping. How the calculation works is something like this: start by defining your "vacuum" as the state in the infinite past infinitely far away from the black hole (or the collapsing star, which I believe is what Hawking actually used) to have no particles in it. Then, propagate this state through (Schwarzschild) time and see what shows up on a surface infinitely far away from the black hole in the infinite future. What one finds is that the state is now populated with a purely thermal distribution of particles. There are ways of doing the calculation with correlation functions of local operators, but this gets very obscure very fast, so I won't get into it.
I think that you could, in principle, view it the way that you want to. There is a conserved current which corresponds to the conservation of energy in this spacetime. We see a positive flux associated with this current at infinity (this is the radiation of positive-frequency particles), so to conserve this current, you could say that there is a negative flux going into the black hole. Another way of saying this is that there is a positive flux of negative energy is going into the black hole. In fact, Wald even words it this way in his book: "By conservation of J^a, a negative energy flux equal in magnitude ... must flow into the black hole." But these are not antiparticles, this is just another way of saying that the energy going out to infinity had to come from somewhere to conserve energy.
I also think I may have worded my previous comment a bit awkwardly. It's best not to think of the energy as coming from the curvature itself, but instead from the source of energy which causes the curvature. So in this case, there is a black hole with mass M which is causing the curvature. As the black hole radiates particles, the only source of energy for these particles to pull from is the mass, so the mass must decrease. In Wald's way of saying it, the negative energy current must enter the black hole, which again just serves to decrease the mass.
This whole question I think starts to get a bit sketchy when talking about charged black holes since all of this is done for non-interacting fields. As soon as we throw in charge, we necessarily introduce particle interactions and the whole calculation essentially goes out the window. I'm not sure whether or not this has been attempted, but I have not ever seen a similar calculation for a black hole emitting charged particles. (Of course, one could have a charged black hole emitting neutral particles and I think the result would be very similar to the uncharged case, with some extra complications).
@@zapphysics Thank you very much, this helped a lot! I hope you had a happy easter weekend.
Maybe I get too hung up on Hawking's analogy, but if he didn't associate a negative energy with the infalling particles, why did he think that analogy was useful at all?
You're not the first person I hear say that the energy comes from the curvature around the horizon. And since that curvature is really what defines the Schwarzschild black hole I don't know if that isn't a useful way to think about it.
To be honest I find it very unsatisfying to say the black hole must lose energy because energy must be conserved and the black hole is the only available source. I don't usually have trouble with abstractions, so I don't think I'm looking for an actual mechanism, but this irks me somehow.
I also read that we can think about the black hole as a giant excited atom which gives off energy as it slowly falls to the ground state(which I suppose would be the vacuum?) and that negative energy partner(which I see in many, many descriptions of Hawking Radiation) is what deexcites it. Does that mean we can model an atom emitting a photon as negative energy flowing from the EM field into the atom? Or am I reading too much into analogies again?
I'd wager there is a mich simpler way to look at the negative energy solutions... Chirality :) instead of reversing momentum, you can use the left hand rule to make the positron move like the "time-reversed" electron. The "hand rules" of the vector product come from the algebra of quaternions and more generally geometric algebra, which has a natural definition of handedness built-in. You could very well solve the - sign problem by introducing an extra parameter £, which takes values of +1 and -1 standing respectively for a left handed and right handed choice of basis vectors. This would let you keep both positive energy and positive momentum in both solutions.
Dirac went nah nah, in reality there exists an infinite sea of negative electrons! Don't be like Dirac.
It is a really awesome video, but every time you say "Schrödinger" with a soft g, my ears bleed
Eavin Schroodingeh